An agricultural manager requires

The bulk modulus of a material is 3.5 ✕ 1011 N/m2. What percent fractional change in volume does a piece of this material underg

kiruha [24]

Answer:

percentage change in volume = 0.00285 %

Explanation:

given data

bulk modulus = 3.5 × $10^{11}$ N/m²

bulk stress = $10^{7}$ N/m²

solution

we will apply here bulk modulus formula that is

bulk modulus = $\frac{bulk\ stress}{bulk\ strain}$ ...............1

put here value and we get

3.5 × $10^{11}$ = $\frac{10^7}{bulk\ strain}$

solve it we get

bulk strain = 2.8571 × $10^{-5}$

and

bulk strain = $\frac{change\ volume}{original\ volume}$

so that percentage change in volume is = 2.8571 × $10^{-5}$ × 100

percentage change in volume = 0.00285 %

6 0

4 years ago

A biotechnology company produced 225 doses of somatropin, including 11 which were defective. Quality control test 15 samples at

Radda [10]

Answer:

0.59

Explanation:

The batch is rejected if any of the random samples are found defective, or, what is the same, it will be accepted only if all 15 samples are good.

The probability that none be defective is the same probability that all the samples are good. Thus, start to calculate the probability that the batch is accepted.

The probability that the first sample is good is 214 /225, because there are 225 - 11 = 214 good samples in 225 doses.

The probability that the second samples is good too is 213/224, because there is 1 less good sample, in the 224 remaining samples.

By the same process, you conclude that the consecutive probabilities of selecting a good sample are: 212/223, 211/222, 210/221, . . . up to 199/211.

The joint probability of all the samples are good is the product of each probability:

$\frac{214}{225}\cdot\frac{213}{224}\cdot\frac{212}{223}\cdot\frac{211}{222}\cdot\frac{210}{221}\cdot\frac{209}{220}\cdot\frac{208}{219}\cdot\frac{207}{218}\cdot\frac{206}{217}\cdot\frac{205}{216}\cdot\frac{204}{215}\cdot\frac{203}{214}\cdot\frac{202}{213}\cdot\frac{201}{212}\cdot\frac{200}{211}\cdot\frac{199}{210}$

The result is: 0.41278 ≈ 0.41

The conclusion is that the probability that all the samples are good and the batch is accepted is 0.41.

Therefore, the probability that the batch is rejected is 1 - 0.41 = 0.59.

4 0

4 years ago

The underground cafe has an operating cash flow of $187,000 and a cash flow to creditors of $71,400 for the past year. During th

Serggg [28]

Answer:

cash flow to stockholders = $39,700

Explanation:

Operating cash flow = $187,000

cash flow to creditors = $71,400

Net working capital = $28,000

Net capital spending = $47,900

Cash flow to stockholders = ?

CFF = operating cash flow - net working capital - net capital spending

CFF = $187,000 - $28,000 - $47,900 = $111,100

CFF = cash flow to creditors + cash flow to stockholders

cash flow to stockholders = CFF - cash flow to creditors

cash flow to stockholders = $111,100 - $71,400 = $39,700

Hence $39,700 is the amount of the cash flow to stockholders for the last year.

3 0

3 years ago

All of these are true about GMA (MIG) welding EXCEPT that:

Hatshy [7]

Answer:

the welding gun liner regulates the shielding gas.

Explanation:

The purpose of the welding gun liner is to properly position the welding wire from the wire feeder till it gets to the nozzle or contact tip of the gun. Regulation of the shielding gas depends on factors such as the speed, current, and type of gas being used. In gas metal arc welding, an electric arc is used to generate heat which melts both the electrode and the workpiece or base metal.

The electric arc produced is shielded from contamination by the shielding gas. The heat generated by the short electric arc is low.

3 0

3 years ago

How high a building could fire hoses effectively spray from the ground? Fire hose pressures are around 1 MPa. (It is also said t

Mrac [35]

Answer:

$z_{2} = 91.640\,m$

Explanation:

The phenomenon can be modelled after the Bernoulli's Principle, in which the sum of heads related to pressure and kinetic energy on ground level is equal to the head related to gravity.

$\frac{P_{1}}{\rho\cdot g} + \frac{v_{1}^{2}}{2\cdot g}= z_{2}+\frac{P_{2}}{\rho\cdot g}$

The velocity of water delivered by the fire hose is:

$v_{1} = \frac{(300\,\frac{gal}{min} )\cdot(\frac{3.785\times 10^{-3}\,m^{3}}{1\,gal} )\cdot(\frac{1\,min}{60\,s} )}{\frac{\pi}{4}\cdot (0.3\,m)^{2}}$

$v_{1} = 0.267\,\frac{m}{s}$

The maximum height is cleared in the Bernoulli's equation:

$z_{2}= \frac{P_{1}-P_{2}}{\rho\cdot g} + \frac{v_{1}^{2}}{2\cdot g}$

$z_{2}= \frac{1\times 10^{6}\,Pa-101.325\times 10^{3}\,Pa}{(1000\,\frac{kg}{m^{3}} )\cdot(9.807\,\frac{m}{s^{2}} )} + \frac{(0.267\,\frac{m}{s} )^{2}}{2\cdot (9.807\,\frac{m}{s^{2}} )}$

$z_{2} = 91.640\,m$

7 0

3 years ago